Calcium crystallizes in a body-centered cubic structure at 467°C. (b) How many nearest neighbors does each Ca atom possess? (c) Estimate the length of the unit cell edge, a, from the atomic radius of calcium (1.97 Å). (d) Estimate the density of Ca metal at this temperature.

Verified step by step guidance

1. To determine the number of nearest neighbors in a body-centered cubic (BCC) structure, recall that each atom in a BCC lattice has 8 nearest neighbors. This is because the central atom in the cube is surrounded by 8 corner atoms.
2. To estimate the length of the unit cell edge, a, in a BCC structure, use the relationship between the atomic radius (r) and the edge length: a = 4r/√3. Substitute the given atomic radius of calcium (1.97 Å) into this formula to find the edge length.
3. To estimate the density of calcium metal, first calculate the volume of the unit cell using the edge length: V = a^3. Ensure that the edge length is converted to centimeters if necessary, as density is typically expressed in g/cm³.
4. Next, determine the mass of the unit cell. In a BCC structure, there are 2 atoms per unit cell. Calculate the mass of these atoms using the molar mass of calcium (40.08 g/mol) and Avogadro's number (6.022 x 10^23 atoms/mol).
5. Finally, calculate the density using the formula: density = mass/volume. Substitute the mass of the unit cell and the volume you calculated to find the density of calcium metal at 467°C.